On 12 Apr., 21:01, Dan <dan.ms.ch...@gmail.com> wrote:> > So do you claim that d is not in the list but all its finite initial> > segments (d_1, ..., d_n) are in the list? If so, what is the> > difference>> Any finite initial segment of any number is a rational number .

Yes. Let us call it FIS.

> Therefore, forall m, the list of rational numbers contains any finite> inital segment of m.> Does that mean that the list of rational numbers contains all> numbers?

No, but it contains all decimal fractions, i.e., all terminatingdecimals. There is no chance to represent 1/3 or sqrt(2) by digits.There is no chance to apply Cantor's argument other than to FISs.Every d_n is part of a FIS.

In order to show that, I'd like to use the Binary Tree.Consider the path 0.111...It is the union of all its FISs0.10.110.111...This is a strictly incresing infinite sequence that does not containits limit 0.111...

However, since every FIS is the union of itself with all itspredecessors, this sequence is also a sequence of unions, and, sinceit is not finite, contains unions of more than n FIS (for every n youcan take). Usually this is called infinite uninon. So, if the path0.111... is the union of all its FIS, then it should also be in list,since the list contains all finite unions, i.e., an infinite union.(Compare the union of all finite natural numbers results in aninfinite set.)

So we obtain a contradiction. 0.111... is in the list, if the list isunderstood as sequence of unions like{1} U {1, 2} U {1, 2, 3} U ... = |N,and it is not in the list, if the list is understood as sequence inmathematics.

Since the Binary Tree deletes the difference between sequence andunions, it shows this problem in full clarity.

The Binary Tree is constructed by countably many nodes. So it isimpossible to distinguish uncountably many paths *by nodes*.